Question

Name the reducing agent in the following:
\[3Mn{O_2} + 4Al \to 3Mn + 2A{l_2}{O_3}\]
\[
  A.Mn \\
  B.Al \\
  C.Mn{O_2} \\
  D.A{l_2}{O_3} \\
 \]

Answer 1

Hint: Try to find out the oxidation states of the compounds for an easy answer. The free elements will have zero oxidation states. The reducing agent will reduce the other compound and itself gets oxidized. So find out which element is going from lower oxidation state to higher oxidation state that will be the reducing agent.

Complete answer:
The reaction is given as: \[3Mn{O_2} + 4Al \to 3Mn + 2A{l_2}{O_3}\]
This reaction is a redox reaction, and so in this reaction one of the elements is undergoing oxidation and the other element is undergoing reduction.
The oxidation and reduction are carried out by the oxidizing and reducing agents. Oxidizing agents are species that tend to oxidize other substances; this means an increase in the oxidation state of the substance by making it lose electrons and itself they get reduced. And reducing agents are species that reduce other substances and they get oxidized.
In this reaction, to find out the oxidizing and reducing species; we will look into the oxidation states of the two species. The oxidation states of the free element will be zero, this means $Al$ and $Mn$will have zero oxidation state. The compounds left are \[Mn{O_2}\]and \[A{l_2}{O_3}\]. The oxidation state of $Mn$ in \[Mn{O_2}\] will be $ + 4$ and of $Al$ in \[A{l_2}{O_3}\] will be $ + 3$.
\[\mathop {3Mn{O_2}}\limits^{ + 4} + 4\mathop {Al}\limits^0 \to 3\mathop {Mn}\limits^0 + \mathop {2A{l_2}{O_3}}\limits^{ + 3} \]
So it can be seen here that $Mn$is going from $ + 4$ to $0$ oxidation state and so it is undergoing reduction, which means that it is an oxidizing agent. And $Al$ is going from $0$ to $ + 3$ oxidation state, and so it is undergoing oxidation which means it is a reducing agent. Here $Mn$ is oxidizing $Al$ to $ + 3$ oxidizing state and $Al$ is reducing $Mn$ to $0$ oxidation state.

Therefore the correct option is B. $Al$

Note:
Oxidation is the process where there is removal of an electron from the molecule. And reduction is the process where there is gain of electrons by the molecule during a chemical reaction. The oxidation states of the two compounds can be found out in the following manner:
${\text{for }}Mn{O_2}$
$ \Rightarrow x + 2( - 2) = 0$
$ \Rightarrow x = + 4$
And ${\text{for }}A{l_2}{O_3}$
$ \Rightarrow 2x + 3( - 2) = 0$
$ \Rightarrow 2x = + 6$
$ \Rightarrow x = + 3$
Therefore oxidation state of $Mn{\text{ in }}Mn{O_2} = + 4$ and $Al{\text{ in }}A{l_2}{O_3} = + 3$